C++ Program To Delete N Nodes After M Nodes Of A Linked List Last Updated : 30 Mar, 2022 Summarize Comments Improve Suggest changes Share Like Article Like Report Given a linked list and two integers M and N. Traverse the linked list such that you retain M nodes then delete next N nodes, continue the same till end of the linked list.Difficulty Level: Rookie Examples: Input: M = 2, N = 2 Linked List: 1->2->3->4->5->6->7->8 Output: Linked List: 1->2->5->6 Input: M = 3, N = 2 Linked List: 1->2->3->4->5->6->7->8->9->10 Output: Linked List: 1->2->3->6->7->8 Input: M = 1, N = 1 Linked List: 1->2->3->4->5->6->7->8->9->10 Output: Linked List: 1->3->5->7->9Recommended: Please solve it on "PRACTICE" first, before moving on to the solution. The main part of the problem is to maintain proper links between nodes, make sure that all corner cases are handled. Following is C implementation of function skipMdeleteN() that skips M nodes and delete N nodes till end of list. It is assumed that M cannot be 0. C++ // C++ program to delete N nodes // after M nodes of a linked list #include <bits/stdc++.h> using namespace std; // A linked list node class Node { public: int data; Node *next; }; /* Function to insert a node at the beginning */ void push(Node ** head_ref, int new_data) { // Allocate node Node* new_node = new Node(); // Put in the data new_node->data = new_data; // Link the old list off the // new node new_node->next = (*head_ref); // Move the head to point to // the new node (*head_ref) = new_node; } // Function to print linked list void printList(Node *head) { Node *temp = head; while (temp != NULL) { cout<<temp->data<<" "; temp = temp->next; } cout<<endl; } // Function to skip M nodes and then // delete N nodes of the linked list. void skipMdeleteN(Node *head, int M, int N) { Node *curr = head, *t; int count; // The main loop that traverses // through the whole list while (curr) { // Skip M nodes for (count = 1; count < M && curr!= NULL; count++) curr = curr->next; // If we reached end of list, // then return if (curr == NULL) return; // Start from next node and delete // N nodes t = curr->next; for (count = 1; count<=N && t!= NULL; count++) { Node *temp = t; t = t->next; free(temp); } // Link the previous list with // remaining nodes curr->next = t; // Set current pointer for next // iteration curr = t; } } // Driver code int main() { /* Create following linked list 1->2->3->4->5->6->7->8->9->10 */ Node* head = NULL; int M=2, N=3; push(&head, 10); push(&head, 9); push(&head, 8); push(&head, 7); push(&head, 6); push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout << "M = " << M<< " N = " << N << "Given Linked list is :"; printList(head); skipMdeleteN(head, M, N); cout<<"Linked list after deletion is :"; printList(head); return 0; } // This code is contributed by rathbhupendra Output: M = 2, N = 3 Given Linked list is : 1 2 3 4 5 6 7 8 9 10 Linked list after deletion is : 1 2 6 7 Time Complexity: O(n) where n is number of nodes in linked list. Auxiliary Space: O(1) Please refer complete article on Delete N nodes after M nodes of a linked list for more details! Comment More infoAdvertise with us Next Article C++ Program To Delete Alternate Nodes Of A Linked List K kartik Follow Improve Article Tags : Linked List C++ Programs DSA Linked Lists Microsoft Amazon +2 More Practice Tags : AmazonMicrosoftLinked List Similar Reads C++ Program To Delete Alternate Nodes Of A Linked List Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3. Recomm 3 min read C++ Program To Delete Middle Of Linked List Given a singly linked list, delete the middle of the linked list. 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